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leetcode刷题日记(303.区域和检索 - 数组不可变)

发表于 2021-03-23 | 最后更新于 2021-12-19 | 开发

给定一个整数数组  nums,求出数组从索引 i 到 j(i ≤ j)范围内元素的总和,包含 i、j 两点。

实现 NumArray 类:

  • NumArray(int[] nums) 使用数组 nums 初始化对象
  • int sumRange(int i, int j) 返回数组 nums 从索引 i 到 j(i ≤ j)范围内元素的总和,包含 i、j 两点(也就是 sum(nums[i], nums[i + 1], ... , nums[j]))  

示例:

输入:
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
输出:
[null, 1, -1, -3]

解释:
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) 
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))

提示:

  • 0 <= nums.length <= 104
  • -105 <= nums[i] <= 105
  • 0 <= i <= j < nums.length
  • 最多调用 104 次 sumRange 方法

@芜桐的解法

class NumArray:

    def __init__(self, nums: List[int]):
        self.nums = nums

    def sumRange(self, left: int, right: int) -> int:
        ans = 0
        while left <= right:
            ans += self.nums[left]
            left += 1
        return ans



# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(left,right)